# 2021 Fall AMC 12A Problems/Problem 11

## Problem

Consider two concentric circles of radius $17$ and $19.$ The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?

$\textbf{(A)}\ 12\sqrt{2} \qquad\textbf{(B)}\ 10\sqrt{3} \qquad\textbf{(C)}\ \sqrt{17 \cdot 19} \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 8\sqrt{6}$

## Solution 1 (Pythagorean Theorem)

Label the center of both circles $O$. Label the chord in the larger circle as $\overline{ABCD}$, where $A$ and $D$ are on the larger circle and $B$ and $C$ are on the smaller circle. Construct the radius perpendicular to the chord and label their intersection as $M$. Because a radius that is perpendicular to a chord bisects the chord, $M$ is the midpoint of the chord.

Construct segments $\overline{AO}$ and $\overline{BO}$. These are radii with lengths 17 and 19 respectively.

Then, use the Pythagorean Theorem. In $\triangle OMA$, we have \begin{align*} OM^2 & = OA^2 - AM^2 \\ & = OA^2 - \left( \frac{AD}{2} \right)^2 \\ & = 19^2 - \frac{AD^2}{4} . \end{align*}

In $\triangle OMB$, we have \begin{align*} OM^2 & = OB^2 - BM^2 \\ & = OB^2 - \left( \frac{BC}{2} \right)^2 \\ & = OB^2 - \left( \frac{AD}{4} \right)^2 \\ & = 17^2 - \frac{AD^2}{16} . \end{align*}

Equating these two expressions, we get $$19^2 - \frac{AD^2}{4} = 17^2 - \frac{AD^2}{16}$$ and $AD=\boxed{\textbf{(E) }8 \sqrt{6}}$.

~eisthefifthletter and Steven Chen

## Solution 2 (Power of a Point)

Draw the diameter perpendicular to the chord. Notice that by symmetry this diameter bisects the chord.

Call the intersection between that diameter and the chord $A$. In the smaller circle, let the shorter piece of the diameter cut by the chord be $x$, making the longer piece $34-x.$ In that same circle, let the $y$ be the length of the portion of the chord that is cut by the diameter.

The radius of the larger circle is $2$ more than the radius of the small circle. So, in the larger circle, the shorter piece of the diameter cut by the chord is of length $x+2$ and the longer piece is $36-x.$ As given in the problem, the bisected length of the chord in the larger circle is twice as much, so it is of length $2y$. By Power of a Point, we can construct a system of equations $$x(34-x) = y^2$$ $$(x+2)(36-x) = (2y)^2.$$ Expanding both equations, we get $$34x-x^2 = y^2$$ $$36x-x^2+72-2x = 4y^2,$$ in which the $34x$ and $-x^2$ terms magically cancel when we subtract the first equation from the second equation. Thus, now we have $$72 = 3y^2 \implies y = 2\sqrt{6} \implies 4y = \boxed{\textbf{(E)} \: 8\sqrt{6}}.$$

-fidgetboss_4000

## Solution 3 (Coordinate Geometry)

Represent the circles as $x^{2}+y^{2}=17^{2}$, and $x^{2}+y^{2}=19^{2}$. Solving for $x$ in these equations we obtain $x=\sqrt{17^{2}-y^{2}}$ and $x=\sqrt{19^{2}-y^{2}}$. Because half of the chord is in the smaller circle, the larger circle should have an $x$ value that is twice as big as the smaller circle's $x$ value. We set up the equation: $$2\sqrt{289-y^{2}}=\sqrt{361-y^{2}}$$ $$4(289-y^{2})=361-y^{2}$$ $$y^{2}=265$$ Substituting $y^{2}$ into $x^{2}+y^{2}=19^{2}$, we obtain $x=\sqrt{96}=4\sqrt{6}$. However, this is only half of the chord length, so we must double it to obtain $\boxed{\textbf{(E)} \: 8\sqrt{6}}.$

-helpmebro

~IceMatrix

## Video Solution (Logic and Geometry)

~Education, the Study of Everything